Question

Let A be the set of all real solutions of equation $x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0$ and B be the set of all real solutions of equation $x^2 - |x| - 12 = 0$ then number of subsets of the set $A \times B$ is

• A. 2
• B. 4
• C. 8
• D. 16

Answer 1

Answer: (B)

4

Answer 2

1. Finding A:

The equation is

$$x\left(x^2 + 3|x| + 5|x-1| + 6|x-2|\right) = 0.$$

This gives either $x = 0$ or

$$x^2 + 3|x| + 5|x-1| + 6|x-2| = 0.$$

Since each term in the sum is nonnegative, the sum is 0 only if every term is 0.

• $x^2 = 0$ implies $x=0$.
• But $|x| = 0$ also forces $x=0$, and $|x-1|=0$ would require $x=1$ (and similarly for $|x-2|$).

Hence, the only solution is $x = 0$.

$\therefore A = \{0\}$.

2. Finding B:

The equation is

$$x^2 - |x| - 12 = 0.$$

Case 1: $x \ge 0$ so $|x| = x$:

$$x^2 - x - 12 = 0 \quad \Rightarrow \quad (x-4)(x+3)=0.$$

$x = 4$ (since $x = -3$ is not valid for $x\ge0$).

Case 2: $x < 0$ so $|x| = -x$:

$$x^2 + x - 12 = 0.$$

Using the quadratic formula,

$$x = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}.$$

This gives $x = 3$ and $x = -4$. Only $x = -4$ satisfies $x < 0$.

Hence, $B = \{-4, 4\}$.

3. Finding number of subsets of $A \times B$:

$$A \times B = \{(0, -4), (0, 4)\}.$$

The number of elements is $2$.

The number of subsets of a set with $n$ elements is $2^n$.

Therefore, the number of subsets is $2^2 = 4$.